Question

The number of integral value(s) of $a$ for which one root of the equation $(a-5)x^2-2ax+a-4=0$ is smaller than $1$ and the other greater than $2,$ is

• A. 18.00
• B. 18.0
• C. 18

Answer 1

Answer: (A), (B), (C)

$(a-5)x^2-2ax+a-4=0$
$(a\ne 5)$ as equation has two roots.
$\Rightarrow x^2-\left(\frac{2a}{a-5}\right)x+\left(\frac{a-4}{a-5}\right)=0$
Let $f\left(x\right)=x^2-\left(\frac{2a}{a-5}\right)x+\left(\frac{a-4}{a-5}\right)$

Required conditions are
$\left(i\right)f\left(1\right)<0$ and $\left(ii\right)f\left(2\right)<0$

$\left(i\right)f\left(1\right)<0$
$\Rightarrow \frac{(-9)}{a-5}<0$
$\Rightarrow a\in (5,\infty )...\left(1\right)$

$\left(ii\right)f\left(2\right)<0$
$\Rightarrow \frac{a-24}{a-5}<0$
$\Rightarrow 5<a<24$
$\Rightarrow a\in (5,24)...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$a\in (5,24)$

Hence, the number of integral values of $a$ is $18.$